Final answer:

PH of acid = -㏒₁₀[H3O⁺]

pH of buffer solution is pH = (- ㏒₁₀Kₐ) + 0.5224

Given that: We are given [NaF] = 0.05 mol, [HF] = 0.015 mol,

[H₂SO] = 0.01 M

To find: We have to find pH of 0.01 M H₂SO, acid, pH of buffer solution NaF and HF

Explanation:

• you have not provided the question correctly. i.e, you have not given acid formula completely and Ka value of HF. you have provided K value which means Kₐ value with power of 10 is + not given K value correctly in the second part of the question. I give answer of that part without putting the value of Kₐ.

• So i'm giving you answer as per my knowledge. May be it is not exactly same question in your book that i will answered here. So next time could you please upload full question correctly as in your book then you will get correct answer.

• pH of a solution is the negative logarithm to the base 10 of concentration of hydronium ion in moles per litre.

                  pH = - ㏒₁₀[H3O⁺]

at 298 K, neutral solution [H3O⁺] = 10⁷ M so pH = 7

in the case of acidic solution [H3O⁺] > 10⁷ M so pH < 7

in the case of basic solution [H3O⁺] < 10⁷ M so pH > 7

• From the equation of pH we have to find [H3O⁺] ion

               [H3O⁺] = [tex]10^{-pH}[/tex]

• Coming to second part of the question, according to Henderson's equation for finding pH of acidic buffer is

                pH = pKa+ ㏒₁₀[[tex]\frac{salt}{acid}[/tex]]

• pKa = -㏒₁₀Kₐ here Ka value not given correctly so I will put it as Kₐ

concentration of salt = [salt] = 0.05 M

concentration of acid = [acid] = 0.015 M

• Put these values in Henderson's equation

          pH = -㏒₁₀ Kₐ + ㏒(0.05/0.015)

                = - ㏒₁₀ Kₐ + ㏒ (3.33)

                = - ㏒₁₀Kₐ + 0.5224

To know more about the concept please go through the links

https://brainly.in/question/1865794

https://brainly.in/question/17617745

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pH of the acid is 2 and pH of the buffer is 1.33.

Given,

Molarity of H₂SO₄= 0.01M

Molarity of salt NaF =0.05M

Molarity of HF= 0.015M

To find,

pH of 0.01 M H₂SO₄

pH of buffer solution

Solution,

pH when molarity is given, can be calculated as:

pH= -log[M]

pH =-log[0.01M]

pH = 2

Now,

According to Henderson's equation for finding pH of acidic buffer:

pH = pKa+ ㏒₁₀[salt /acid]

pKa = -log Ka

pKa = -log[7.2*10]

pKa = -1.857

pH = pKa+ ㏒₁₀[salt /acid]

pH = -1.857 + log[0.05/0.015]

pH =1.33

Hence, pH of acid is 2 and pH of buffer is 1.33.

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